/*
 * @lc app=leetcode.cn id=257 lang=cpp
 *
 * [257] 二叉树的所有路径
 *
 * https://leetcode.cn/problems/binary-tree-paths/description/
 *
 * algorithms
 * Easy (69.79%)
 * Likes:    761
 * Dislikes: 0
 * Total Accepted:    219.2K
 * Total Submissions: 313.9K
 * Testcase Example:  '[1,2,3,null,5]'
 *
 * 给你一个二叉树的根节点 root ，按 任意顺序 ，返回所有从根节点到叶子节点的路径。
 *
 * 叶子节点 是指没有子节点的节点。
 *
 *
 * 示例 1：
 *
 *
 * 输入：root = [1,2,3,null,5]
 * 输出：["1->2->5","1->3"]
 *
 *
 * 示例 2：
 *
 *
 * 输入：root = [1]
 * 输出：["1"]
 *
 *
 *
 *
 * 提示：
 *
 *
 * 树中节点的数目在范围 [1, 100] 内
 * -100 <= Node.val <= 100
 *
 *
 */

// @lc code=start
#include <iostream>
#include <vector>
using namespace std;
// Definition for a binary tree node.
/* struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
}; */

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode *root) {
        vector<string> paths;
        dfs(root, "", paths);
        return paths;
    }

    // helper
    void dfs(TreeNode *node, string path, vector<string> &paths) {
        if (node == nullptr) {
            return;
        }
        path += to_string(node->val);
        if (node->left == nullptr && node->right == nullptr) {
            paths.push_back(path);
        }
        else {
            path += "->";
            dfs(node->left, path, paths);
            dfs(node->right, path, paths);
        }
    }
};
// @lc code=end
